Chapter 8, Problem 85AP

To determine

The centre of gravity of the gymnast for a given segment.

Answer to Problem 85AP

Answer The centre of gravity of gymnast is $0.460\text{m,}0.0955\text{m}$ for x and y values respectively.

Explanation of Solution

Explanation

Given Info:

The mass of the arms is $6.87\text{kg}$ , the mass of torso is $33.57\text{kg}$ , the mass of thighs is $14.07\text{kg}$ , the mass of legs is $7.54\text{kg}$

From the table the length of arms is $0.548\text{m}$ , the length of torso is $0.601\text{m}$ , the length of thighs is $0.374\text{m}$ , the length of legs is $0.350\text{m}$

From the table the centre of gravity of arms is $0.239\text{m}$ along y direction, the centre of gravity of torso $0.337\text{m}$ along x direction, the centre of gravity of thighs is  $0.151\text{m}$ , the centre of gravity of legs is $0.227\text{m}$. The angle between thigh and horizontal is $60°$.

Formula to calculate the position of centre of gravity of gymnast along x-axis is,

$x=\frac{m_{\text{a}}{x}_{\text{a(cg)}}+m_t{x}_{\text{t(cg)}}+m_{\text{th}}{x}_{\text{th(cg)}}+m_{\text{l}}{x}_{\text{l(cg)}}}{m_{\text{a}}+m_t+m_{\text{th}}+m_{\text{l}}}$

• x is the centre of gravity of gymnast along x axis
• $m_{\text{a}}$ is the mass of arms
• $m_{\text{t}}$ is the mass of torso
• $m_{\text{th}}$ is the mass of thighs
• $m_{\text{l}}$ is the mass of legs
• $x_{\text{a(cg)}}$ is the centre of mass in the x direction of the arms
• $x_{\text{t(cg)}}$ is the centre of mass in the x direction of the torso
• $x_{\text{th(cg)}}$ is the centre of mass in the x direction of the thighs
• $x_{\text{l(cg)}}$ is the centre of mass in the x direction of the legs

Formula to calculate the position of centre of gravity of gymnast along y-axis is,

$y=\frac{m_{\text{a}}{y}_{\text{a(cg)}}+m_t{y}_{\text{t(cg)}}+m_{\text{th}}{y}_{\text{th(cg)}}+m_{\text{l}}{y}_{\text{l(cg)}}}{m_{\text{a}}+m_t+m_{\text{th}}+m_{\text{l}}}$

• y is the centre of gravity of gymnast along y axis.
• $y_{\text{a(cg)}}$ is the centre of mass in the y direction of the arms
• $y_{\text{t(cg)}}$ is the centre of mass in the y direction of the torso
• $y_{\text{th(cg)}}$ is the centre of mass in the y direction of the thighs
• $y_{\text{l(cg)}}$ is the centre of mass in the y direction of the legs

Formula to calculate the centre of gravity of thighs along x direction is,

$x_{\text{th(cg)}}=L_{\text{t}}+r_{\text{cg(th)}}\cos \theta$

• $L_{\text{t}}$ is the length of the torso
• $r_{cg\text{(th)}}$ is the centre of gravity of thighs
• $\theta$ is the angle between thighs and horizontal x axis

Substitute $0.601\text{m}$ for $L_{\text{t}}$ , $0.151\text{m}$ for $r_{cg\text{(th)}}$, and $60°$ for $\theta$ to calculate $x_{\text{th(cg)}}$.

$\begin{array}{c}{x}_{\text{th(cg)}}=0.601\text{m}+\left(0.151\text{m}\right)\cos 60°\\ =0.601\text{m}+\left(0.151\text{m}\right)\frac{1}{2}\\ =0.677\text{m}\end{array}$

Formula to calculate the centre of gravity of thighs along y direction is,

$y_{\text{th(cg)}}=r_{\text{cg(th)}}\sin \theta$

Substitute $0.151\text{m}$ for $r_{cg\text{(th)}}$ and $60°$ for $\theta$ to calculate $y_{\text{th(cg)}}$

$\begin{array}{c}{y}_{\text{th(cg)}}=\left(0.151\text{m}\right)\sin 60°\\ =\left(0.151\text{m}\right)\frac{\sqrt{3}}{2}\\ =0.131\text{m}\end{array}$

Formula to calculate the centre of gravity of legs along x direction is,

$x_{\text{l(cg)}}=L_{\text{t}}+L_{\text{th}}\cos \theta +r_{\text{cg(l)}}$

• $L_{\text{t}}$ is the length of the torso
• $r_{cg\text{(l)}}$ is the centre of gravity of legs
• $L_{\text{th}}$ is the length of thighs.

Substitute $0.601\text{m}$ for $L_{\text{t}}$ , $0.227\text{m}$ for $r_{cg\text{(l)}}$ , $0.374\text{m}$ for $L_{\text{th}}$ and $60°$ for $\theta$ to calculate $x_{\text{l(cg)}}$

$\begin{array}{c}{x}_{\text{l(cg)}}=\left[0.601\text{m}+\left(0.374\text{m}\right)\cos 60°+0.227\text{m}\right]\\ =\left[0.601\text{m}+\left(0.374\text{m}\right)\frac{1}{2}+0.227\text{m}\right]\\ =\text{1}\text{.02}\text{m}\end{array}$

Formula to calculate the centre of gravity of legs along y direction is,

$y_{\text{l(cg)}}=L_{t\text{h}}\sin \theta$

Substitute $0.374\text{m}$ for $r_{cg\text{(th)}}$ and $60°$ for $\theta$ to calculate $y_{\text{l(cg)}}$

$\begin{array}{c}{y}_{\text{l(cg)}}=\left(0.374\text{m}\right)\sin 60°\\ =\left(0.374\text{m}\right)\frac{\sqrt{3}}{2}\\ =0.324\text{m}\end{array}$

Conclusion:

Formula to calculate the position of centre of gravity of gymnast along x-axis is,

$x=\frac{m_{\text{a}}{x}_{\text{a(cg)}}+m_t{x}_{\text{t(cg)}}+m_{\text{th}}{x}_{\text{th(cg)}}+m_{\text{l}}{x}_{\text{l(cg)}}}{m_{\text{a}}+m_t+m_{\text{th}}+m_{\text{l}}}$

Substitute $6.87\text{kg}$ for $m_{\text{a}}$ , 0 m for $x_{\text{a(cg)}}$, $33.57\text{kg}$ for $m_{\text{t}}$, $0.337\text{m}$ for $x_{\text{t(cg)}}$, $14.07\text{kg}$ for $m_{\text{th}}$ $0.677\text{m}$ for $x_{\text{th(cg)}}$ , $7.54\text{kg}$ for $m_{\text{l}}$ , and $1.02\text{m}$ for $x_{\text{l(cg)}}$ to calculate  x .

$\begin{array}{c}x=\frac{\left(6.87\text{kg}\right)\left(0\right)+\left(33.57\text{kg}\right)\left(0.337\text{m}\right)+\left(14.07\text{kg}\right)\left(0.677\text{m}\right)+\left(7.54\text{kg}\right)\left(1.02\text{m}\right)}{\left(6.87\text{kg}+33.57\text{kg}+14.07\text{kg}+7.54\text{kg}\right)}\\ =\frac{28.53}{62.05}\\ =0.460\text{m}\end{array}$

Formula to calculate the position of centre of gravity of gymnast along y-axis is,

$y=\frac{m_{\text{a}}{y}_{\text{a(cg)}}+m_t{y}_{\text{t(cg)}}+m_{\text{th}}{y}_{\text{th(cg)}}+m_{\text{l}}{y}_{\text{l(cg)}}}{m_{\text{a}}+m_t+m_{\text{th}}+m_{\text{l}}}$

Substitute $6.87\text{kg}$ for $m_{\text{a}}$ ,  $33.57\text{kg}$ for $m_{\text{t}}$, $14.07\text{kg}$ for $m_{\text{th}}$ , $7.54\text{kg}$ for $m_{\text{l}}$ , $0.239\text{m}$ for $y_{\text{a(cg)}}$, 0 m for $y_{\text{t(cg)}}$, $0.131\text{m}$ for $y_{\text{th(cg)}}$ and $0.324\text{m}$ for $y_{\text{l(cg)}}$ to calculate  the centre of  gravity of the gymnast

$\begin{array}{c}y=\frac{\left(6.87\text{kg}\right)\left(0.239\text{m}\right)+\left(33.57\text{kg}\right)\left(0\right)+\left(14.07\text{kg}\right)\left(0.131\text{m}\right)+\left(7.54\text{kg}\right)\left(0.324\text{m}\right)}{\left(6.87\text{kg}+33.57\text{kg}+14.07\text{kg}+7.54\text{kg}\right)}\\ =\frac{5.93}{62.05}\\ =0.0955\text{m}\end{array}$

Therefore the centre of gravity of the gymnast is $0.460\text{m,}0.0955\text{m}$ respectively x and y values